Timeline for Why can't I get global existence to linear PDE in this way?
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8 events
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| Jan 24, 2014 at 15:59 | comment | added | michael_carbon | ..But we get this for the PDE in the OP easily because of the simple nature of the PDE if I am right. | |
| Jan 24, 2014 at 15:59 | comment | added | michael_carbon | I think I finally understand, let me just go back to the PDE I have: it is true that for each $T$ there is a unique solution on $[0,T]$. Defining $v$ the way I did in the OP, we just need to check that $v \in L^2(0,\infty;V)$, which you say holds if the integrability statement you stated is true. I was reading some papers and got confused; I guess in nonlinear PDEs it is in general NOT immediate that "for each $T$ there is a solution on $[0,T]$"; rather there is some $T$ (not given a priori) on which the solution exists. then the task is to show that this $T$ can be extended... | |
| Jan 24, 2014 at 13:37 | comment | added | username | @michael_carbon You SE thread is still about PDE, but I don't think this is what is stopping you. What you want is a proof of the integrability statement I stated, which you should ask on SE. 'My' f is your v, of course. | |
| Jan 24, 2014 at 13:14 | comment | added | michael_carbon | So if I understand you right, it is enough that the data $f$ be uniformly bounded independent of $T$. | |
| Jan 24, 2014 at 13:12 | comment | added | michael_carbon | Thanks for answering. If you (or anyone else) cares to answer I've opened a thread of SE math.stackexchange.com/questions/648887/…). | |
| Jan 23, 2014 at 15:40 | history | edited | username | CC BY-SA 3.0 |
added 114 characters in body
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| Jan 23, 2014 at 15:19 | history | edited | username | CC BY-SA 3.0 |
clarified the answer
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| Jan 23, 2014 at 11:50 | history | answered | username | CC BY-SA 3.0 |